calculusdefinite integralsintegrationmultivariable-calculus
I am facing problem in calculating $$I_1=\int_{[0,1]} \int_{[0,1]} \frac {x^2-y^2}{(x^2+y^2)^2} \,dx\,dy$$ and $$I_2=\int_{[0,1]} \int_{[0,1]} \frac {x^2-y^2}{(x^2+y^2)^2} \,dy\,dx$$ after substituting $x=r\cos \theta$ and $y=r\sin \theta$ we have $I_1=\int_{[-\pi,\pi]} \int_{[0,1]}\frac{\cos(2\theta)}{r^2}r\,dr\, d\theta$ so I am getting $\infty \times 0=0$ So I am not getting $\pi/4$ and $-\pi/4$ resp.
Maybe there is a silly point I am missing. Please help.
Instead of doing the $u$ substitution before evaluating the integral in $v$, it's easier to do it after computing that integral. Here is a detailed computation.
(I've assumed that you want to evaluate this integral and not simply to show that it is equal to $\zeta(2)$. This exercise appears in several references: Tom Apostol's A Proof that Euler Missed: Evaluating $\zeta(2)$ the Easy Way, Martin Aigner and Günter Ziegler's Proofs from The BOOK and as an exercise in a number theory text by LeVeque .)
By the substitution $x=\frac{\sqrt{2}}{2}\left( u-v\right) ,y=\frac{\sqrt{2}}{2}\left( u+v\right) $, whose Jacobian $J=\frac{\partial (x,y)}{\partial (u,v)}=1$, the region of integration becomes the blue square in the $u,v$-plane with vertices
$$(0,0),\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right),(\sqrt{2},0),\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right), $$ as shown in the following figure.
Observing that
$$ \frac{1}{1-xy}=\frac{2}{2-u^2+v^2} $$
is symmetric in $v$, we get
$$\begin{eqnarray*} I &=&\int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy & x=\frac{\sqrt{2}}{2} (u-v) ,\quad y=\frac{\sqrt{2}}{2} (u+v) \\ &=&2\int_{u=0}^{\sqrt{2}/2}\int_{v=0}^{u} \frac{2}{2-u^{2}+v^{2}}\times 1\,du\,dv \\ &&{}+2\int_{u=\sqrt{2}/2}^{\sqrt{2}}\int_{v=0}^{\sqrt{2}-u}\frac{2}{ 2-u^2+v^2}\times 1\,du\,dv \\ &=&4\int_0^{\sqrt{2}/2}\left( \int_{0}^{u}\frac{dv}{2-u^{2}+v^{2}}\right) \,du\, \\ &&{}+4\int_{\sqrt{2}/2}^{\sqrt{2}}\left( \int_{0}^{\sqrt{2}-u}\frac{dv}{ 2-u^{2}+v^{2}}\right) \,du \\ &=&4\int_{0}^{\sqrt{2}/2}\frac{1}{\sqrt{2-u^{2}}}\arctan \frac{u}{\sqrt{ 2-u^{2}}}\,du, & u=\sqrt{2}\sin t \\ &&{}+4\int_{\sqrt{2}/2}^{\sqrt{2}}\frac{1}{\sqrt{2-u^{2}}}\arctan \frac{\sqrt{2 }-u}{\sqrt{2-u^{2}}}\,du\, & u=\sqrt{2}\cos \theta \\ &=&4\int_{0}^{\pi /6}\arctan(\tan t)\, dt\\&&{}+4\int_0^{\pi /3}\arctan \left( \frac{1-\cos \theta}{\sin \theta}\right) d\theta, \\ &=&4\int_{0}^{\pi /6}t\,dt+4\int_0^{\pi /3}\arctan \left( \tan \frac{\theta}{2} \right) \, d\theta \\ &=&\frac{\pi^2}{18}+\frac{\pi^2}{9}=\frac{\pi^2}{6}. \end{eqnarray*} $$
One of the substitutions is a new one. After the first pair of substitutions $x,y$, we have done two additional ones in the resulting integrals, as indicated above: in the 1st, $u=\sqrt{2}\sin t,du=\sqrt{2}\cos t\,dt$, and in the 2nd, $u=\sqrt{2}\cos \theta,du=-\sqrt{2}\sin \theta\,d\theta$.
Let $f(s)$ be defined by
$$ f(s) = \int_{0}^{\infty} \frac{1-\cos x}{x^s} \, \mathrm{d}x $$
for $1 < s < 3$. By writing $\frac{1}{x^s} = \frac{1}{\Gamma(s)} \int_{0}^{\infty} t^{s-1}e^{-xt} \, \mathrm{d}t$, we get
\begin{align*} f(s) &= \int_{0}^{\infty} \frac{t^{s-1}}{\Gamma(s)} \int_{0}^{\infty} (1 - \cos x)e^{-xt} \, \mathrm{d}x \mathrm{d}t \\ &= \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-2}}{1+t^2} \,\mathrm{d}t \\ &= \frac{1}{2\Gamma(s)} \int_{0}^{\infty} \frac{u^{(s-3)/2}}{1+u} \,\mathrm{d}u \qquad (u = t^2) \\ &= \frac{1}{2\Gamma(s)} \cdot \frac{\Gamma\big(\frac{s-1}{2}\big)\Gamma\big(\frac{3-s}{2}\big)}{\Gamma(1)} \\ &= -\frac{\pi}{2\Gamma(s)\cos(\pi s/2)}. \end{align*}
Substituting $x \mapsto \alpha x$ for $\alpha > 0$, this gives
$$ \int_{0}^{\infty} \frac{1-\cos (\alpha x)}{x^s} \, \mathrm{d}x = -\frac{\pi \alpha^{s-1}}{2\Gamma(s)\cos(\pi s/2)} $$
Differentiating $f$ in two ways, one using Leibniz's integral rule and the other using the above formula, we get
$$ \int_{0}^{\infty} \frac{1-\cos(\alpha x)}{x^s}\log x \, \mathrm{d}x = -\frac{\pi\alpha^{s-1}}{2\Gamma(s)\cos(\pi s/2)}\left(\psi(s) + \frac{\pi}{2}\tan(\pi s/2) - \log \alpha\right) $$
and in particular,
$$ \int_{0}^{\infty} \frac{1-\cos(\alpha x)}{x^2}\log x \, \mathrm{d}x = \frac{\pi\alpha}{2}\left(1 - \gamma - \log \alpha\right). $$
Together with OP's computation, this immediately gives the closed form of $I$.
Best Answer
Hint:
$$\frac {x^2-y^2}{(x^2+y^2)^2} = \frac{\partial}{\partial y} \left(\frac{y}{x^2 + y^2}\right) = -\frac{\partial}{\partial x} \left(\frac{x}{x^2 + y^2}\right) $$